Energy time uncertainty principle

Consider a laser source of frequency f. This source is projected to a target at a distance D, so that the light needs time T = D/C to reach the target. I will consider the particle behavior of light in this situation. I will study the motion of one of the photons.

You have to clear on what exactly the energy-time uncertainty principle refers to. ΔE is the uncertainty of the energy distribution of the system, while Δt is the time approximately needed for the mean value of the energy to vary by ΔE. In your example, this principle cannot stand, because the photon has definite energy (i.e. ΔE = 0) and it is kept constant (Δt = infite), so you have the non-definite term ΔE Δt = zero x infinity.

You have to clear on what exactly the energy-time uncertainty principle refers to. ΔE is the uncertainty of the energy distribution of the system, while Δt is the time approximately needed for the mean value of the energy to vary by ΔE. In your example, this principle cannot stand, because the photon has definite energy (i.e. ΔE = 0) and it is kept constant (Δt = infite), so you have the non-definite term ΔE Δt = zero x infinity.

Thanks for your reply.

1) Why Δt is infinite?

2) I have read that uncertainty principle says that its impossible do determine the energy and the time accurately. In the example I provided, it seems that both time and energy are determined accurately.

2) I have read that uncertainty principle says that its impossible do determine the energy and the time accurately. In the example I provided, it seems that both time and energy are determined accurately.

1) Actually, Δt its self in not definite: rate of change = zero , because the energy is constant, therefore, Δt = (uncertainty)/(rate of change) = 0/0 -> not definite (if you considered that ΔE is small but finite, THEN Δt would be infinite)

2) This is a naive statement of the principle but, unfortunately, a famous one. First of all, there is no such observable as “time” . Therefore, Δt cannot stand for the “uncertainty of measuring time”, this is a non-sense expression. I recommend you to prove the principle (it’s not very difficult) in order to full understand what it means.

1) Actually, Δt its self in not definite: rate of change = zero , because the energy is constant, therefore, Δt = (uncertainty)/(rate of change) = 0/0 -> not definite (if you considered that ΔE is small but finite, THEN Δt would be infinite)

2) This is a naive statement of the principle but, unfortunately, a famous one. First of all, there is no such observable as “time” . Therefore, Δt cannot stand for the “uncertainty of measuring time”, this is a non-sense expression. I recommend you to prove the principle (it’s not very difficult) in order to full understand what it means.

A better way to put it : T is not a dynamical variable.

One must remember what the \delta T and the \delta E mean.

It is usually stated "The time-energy uncertainty principle a statement about how the statistical uncertainty in the energy controls the time scale for a change in the system."

When you started the laser, look at the first say 5 nanoseconds, (\delta T = 5 ns) \delta E is clearly non zero.

Can you please give me a link to a simple clear proof? All what I found contain much of ambiguity that I can't understand.

I don’t have in mind a link, so I will show you a simple proof: consider an observable A and the corresponding operator [tex]\hat{A}[/tex] The expectation value of A will be: [tex]\left\langle A \right\rangle =\left( \psi ,\hat{A}\psi \right) [/tex] Assuming that the operator is time-independent, the rate of change of the mean value will be:
[tex] \frac{d}{dt}\left\langle A \right\rangle =\frac{d}{dt}\left( \psi ,\hat{A}\psi \right)=\left( \frac{\partial \psi }{\partial t},\hat{A}\psi \right)+\left( \psi ,\hat{A}\frac{\partial \psi }{\partial t} \right)[/tex]
(FedEx, that’s what I mean by rate of change)and if you use Schrodinger’s equation [tex]\frac{\partial \psi }{\partial t}=\frac{1}{i\hbar }\hat{H}\psi [/tex] in the above expression, you get:
[tex] \frac{d}{dt}\left\langle A \right\rangle =\left( \frac{1}{i\hbar }\hat{H}\psi ,\hat{A}\psi \right)+\left( \psi ,\hat{A}\frac{1}{i\hbar }\hat{H}\psi \right)=-\frac{1}{i\hbar }\left( \psi ,\hat{H}\hat{A}\psi \right)+\frac{1}{i\hbar }\left( \psi ,\hat{A}\hat{H}\psi \right)=\frac{1}{i\hbar }\left( \psi ,\left[ \hat{A},\hat{H} \right]\psi \right)=\frac{1}{i\hbar }\left\langle \left[ \hat{A},\hat{H} \right] \right\rangle[/tex]
Now, you can define a characteristic time scale by which “the mean value of the observable A varies by ΔA, assuming that the rate of change is constant” . I will call this characteristic time scale τ and not Δt, in order to avoid the naïve interpretation of Δt. So:
[tex] \tau \equiv \frac{\Delta A}{\left| d\left\langle A \right\rangle /dt\ \right|}=\hbar \frac{\Delta A}{\left| \left\langle \left[ \hat{A},\hat{H} \right] \right\rangle \right|}[/tex]
Now, let’s take as granted the general uncertainty principle:
[tex] \Delta A\Delta B\ge \frac{1}{2}\left| \left\langle \left[ \hat{A},\hat{B} \right] \right\rangle \right|[/tex]
Set in this expression [tex] \hat{B}=\hat{H} [/tex] in order to get:
[tex]\Delta A\Delta E\ge \frac{1}{2}\left| \left\langle \left[ \hat{A},\hat{H} \right] \right\rangle \right|[/tex]
and according to the definition of τ, from the last expression we get:
[tex] \Delta A\Delta E\ge \frac{\hbar }{2}\frac{\Delta A}{\tau }\Rightarrow \tau \,\Delta E\ge \frac{\hbar }{2} [/tex]
That is the time-energy uncertainty relation. I hope from the derivation of it, you clear it’s meaning.

Thank you very much. Although I admit that I am not qualified enough to understand every step in the derivation, but I can see that this principle is based on the probabilistic point of view of quantum mechanics, that is there is uncertainty arose from this point of view regardless the measuring techniques. I also can see that the time is not an instant with a certain uncertainty, but it is a duration.

I still have two questions regardless the initial situation I have introduced:

1) Is it true that the calculated energy E = hf is nothing more than the most probable value, and so there is uncertainty associated with it?
2) How should we take the time? I assume it is the time needed by the light to travel from the source to the target. Is that true?

1) If the photon has some definite energy, it will also have some definite momentum. But a wavefunction with definite momentum is no localized in space (it is a plane wave), so the photon can be anywhere. In order to localize as much as possible the wavefunction, the distribution of energy has to be dispersed. So there will be some uncertainty…

2) Time is just a parameter that dictates the evolution of the system. Since we are dealing with quantum mechanics (i.e. with probabilities) you should ask the question "if photon's wavefunction at t=ti (the time of emission) gives probability = 1 to be at some place (the source), then what is the time tf for which the wavefunction gives probability = 1 at some other place (the target). The difference between tf and ti is the time needed for the photon to travel from the source to the target.